Question

Suppose a certain planet had two moons, one of which was twice as far from the planet as the other. Which moon would complete one revolution of the planet first?

Answer 1

The innermost moon.

Explanation:

Assuming that the moons orbit the planet in elliptical orbits, then Kepler's third law gives the relationship between the orbital period T and the semi major axis A as `T^2/A^3=`constant.

If the orbital period and semi major axis for the inner moon are `T_1` and `A_1` and for the outer moon are `T_2` and `A_2`. Then `T_1^2/A_1^3=T_2^2/A_2^3`. The outer moon is twice the distance which means that `A_2=2A_1`. Substituting gives `T_1^2/A_1^3=T_2^2/(8A_1^3`. Multiply by `8A_1^3` gives `T_2^2=8T_1^2`. Taking the square root gives `T_2=2sqrt(2)T_1`.

This means that a moon which orbits twice the distance away from the planet takes approximately 2.828 times longer to complete an orbit.